# Minimum characters required to be removed to sort binary string in ascending order – Set 2

• Last Updated : 20 Dec, 2021

Given binary string str of size N, the task is to remove the minimum number of characters from the given binary string such that the characters in the remaining string are in sorted order.

Examples:

Input: str = “1000101”
Output: 2
Explanation: Removal of the first two occurrences of ‘1’ modifies the string to “00001”, which is a sorted order. The string can also be made “00011” by performing 2 removals. Therefore, the minimum count of characters to be removed is 2.

Input: str = “001111”
Output: 0
Explanation: The string is already sorted. Therefore, the minimum count of character to be removed is 0.

Last Occurrence Approach: The optimized approach in linear time and constant space is discussed in Set 1 of this article. Here, we are discussing the Dynamic Programming Approach.

Dynamic Programming Approach: This problem can be solved using dynamic programming by observing the following facts, that if K deletion is required to make the string sorted till ith index and

• Case1: S[i+1] = 1 then minimum number of deletions required to make string sorted till (i+1)th index will also be K as appending 1 to a sorted string will keep string sorted so no more deletion required.
• Case2: S[i + 1] = 0 then we have two way to make string sorted till (i+1)th index that are
• either delete all 1’s before (i+1)th index, or
• delete current 0.

Minimum number of deletion to make string valid till (i+1)th index will be minimum of (numbers of 1’s before (i+1)th index , K+1).

Follow the steps below to solve the problem:

• Initialize the variables count1 as 0 and N is the length of the string s.
• Initialize the vector dp[n+1] with values 0.
• Iterate over the range [0, n) using the variable i and perform the following tasks:
• If s[i] equals 0, then set dp[i+1] as the minimum of count1 or 1 + dp[i].
• Else, set dp[i+1] as dp[i] and increase the value of count1 by 1.
• After performing the above steps, print the value of dp[n] as the answer.

Below is the implementation of the above approach:

## C++

 `// C++ program for the above approach``#include ``using` `namespace` `std;` `// Function to find the minimum number``// of deletions to make``// the string sorted``int` `minDeletions(string s)``{``    ``int` `n = s.size();` `    ``// dp[i+1] stores minimum number of``    ``// deletion to make``    ``// substring(0, i) valid``    ``vector<``int``> dp(n + 1, 0);``    ``int` `count1 = 0;` `    ``for` `(``int` `i = 0; i < n; i++) {``        ``if` `(s[i] == ``'0'``) {` `            ``// Case 1: remove current 0``            ``// Case 2: keep current 0``            ``// then delete all 1 before it``            ``dp[i + 1] = min(count1,``                            ``1 + dp[i]);``        ``}``        ``else` `{` `            ``// Appending 1 is always valid``            ``// if substring(0, i) is sorted``            ``dp[i + 1] = dp[i];``            ``count1++;``        ``}``    ``}``    ``return` `dp[n];``}` `// Driver Code``int` `main()``{``    ``string s = ``"00101101"``;``    ``cout << minDeletions(s);``    ``return` `0;``}`

## Java

 `// Java program for the above approach``import` `java.util.*;``public` `class` `GFG``{``  ` `// Function to find the minimum number``// of deletions to make``// the string sorted``static` `int` `minDeletions(String s)``{``    ``int` `n = s.length();` `    ``// dp[i+1] stores minimum number of``    ``// deletion to make``    ``// substring(0, i) valid``    ``int` `[]dp = ``new` `int``[n + ``1``];``    ``for``(``int` `i = ``0``; i < n + ``1``; i++) {``        ``dp[i] = ``0``;``    ``}``    ``int` `count1 = ``0``;` `    ``for` `(``int` `i = ``0``; i < n; i++) {``        ``if` `(s.charAt(i) == ``'0'``) {` `            ``// Case 1: remove current 0``            ``// Case 2: keep current 0``            ``// then delete all 1 before it``            ``dp[i + ``1``] = Math.min(count1,``                            ``1` `+ dp[i]);``        ``}``        ``else` `{` `            ``// Appending 1 is always valid``            ``// if substring(0, i) is sorted``            ``dp[i + ``1``] = dp[i];``            ``count1++;``        ``}``    ``}``    ``return` `dp[n];``}` `// Driver Code``public` `static` `void` `main(String args[])``{``    ``String s = ``"00101101"``;``    ``System.out.println(minDeletions(s));` `}``}` `// This code is contributed by Samim Hossain Mondal.`

## Python3

 `# Python code for the above approach` `# Function to find the minimum number``# of deletions to make``# the string sorted``def` `minDeletions(s):``    ``n ``=` `len``(s)` `    ``# dp[i+1] stores minimum number of``    ``# deletion to make``    ``# substring(0, i) valid``    ``dp ``=` `[``0``] ``*` `(n ``+` `1``)``    ``count1 ``=` `0` `    ``for` `i ``in` `range``(n):``        ``if` `(s[i] ``=``=` `'0'``):` `            ``# Case 1: remove current 0``            ``# Case 2: keep current 0``            ``# then delete all 1 before it``            ``dp[i ``+` `1``] ``=` `min``(count1, ``1` `+` `dp[i])``        ``else``:` `            ``# Appending 1 is always valid``            ``# if substring(0, i) is sorted``            ``dp[i ``+` `1``] ``=` `dp[i]``            ``count1 ``+``=` `1``    ``return` `dp[n]` `# Driver Code``s ``=` `"00101101"``print``(minDeletions(s))` `# This code is contributed by Saurabh Jaiswal`

## C#

 `// C# program for the above approach``using` `System;``public` `class` `GFG``{` `    ``// Function to find the minimum number``    ``// of deletions to make``    ``// the string sorted``    ``static` `int` `minDeletions(``string` `s)``    ``{``        ``int` `n = s.Length;` `        ``// dp[i+1] stores minimum number of``        ``// deletion to make``        ``// substring(0, i) valid``        ``int``[] dp = ``new` `int``[n + 1];``        ``for` `(``int` `i = 0; i < n + 1; i++)``        ``{``            ``dp[i] = 0;``        ``}``        ``int` `count1 = 0;` `        ``for` `(``int` `i = 0; i < n; i++)``        ``{``            ``if` `(s[i] == ``'0'``)``            ``{` `                ``// Case 1: remove current 0``                ``// Case 2: keep current 0``                ``// then delete all 1 before it``                ``dp[i + 1] = Math.Min(count1,``                                ``1 + dp[i]);``            ``}``            ``else``            ``{` `                ``// Appending 1 is always valid``                ``// if substring(0, i) is sorted``                ``dp[i + 1] = dp[i];``                ``count1++;``            ``}``        ``}``        ``return` `dp[n];``    ``}` `    ``// Driver Code``    ``public` `static` `void` `Main()``    ``{``        ``string` `s = ``"00101101"``;``        ``Console.Write(minDeletions(s));` `    ``}``}` `// This code is contributed by Saurabh Jaiswal`

## Javascript

 ``

Output
`2`

Time Complexity:  O(N)
Auxiliary Space: O(N)

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